Difference between revisions of "Talk:2008 IMO Problems/Problem 4"
(New page: There are many interesting properties fo <math>f</math> on can prove. The most interesting is probably <cmath> f(xy) = f(x)f(y)\ \forall_{x,y \in \mathbb{R}^+}</cmath> So f must be an au...) |
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| − | There are many interesting properties | + | There are many interesting properties of <math>f</math> one can prove. The most interesting could be |
<cmath> f(xy) = f(x)f(y)\ \forall_{x,y \in \mathbb{R}^+}</cmath> | <cmath> f(xy) = f(x)f(y)\ \forall_{x,y \in \mathbb{R}^+}</cmath> | ||
Latest revision as of 10:25, 23 August 2008
There are many interesting properties of 
one can prove. The most interesting could be
![\[f(xy) = f(x)f(y)\ \forall_{x,y \in \mathbb{R}^+}\]](http://latex.artofproblemsolving.com/a/f/3/af3252e23c66759fb0082bbad52f61d67f63f952.png)
So f must be an automorphism of the group 
.
If you assume (or can prove) that f must be continuous, then must be of the form 
for 
. (the only continuous automorphisms of the group)
Plugging this into the original functional equation gives 
.
I would like to see an alternative solution along these lines.
